JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 2)
Imagine that the electron in a hydrogen atom is replaced by a muon ($$\mu$$). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :
13.6 eV
2815.2 eV
331.2 eV
27.2 eV
Explanation
mm = 207 me
In hydrogen atom one electron present. Now that the electron in hydrogen atom is replaced by a muon ($$\mu$$).
We know, Energy(E) = $$ - {{{e^4}m} \over {8\varepsilon _0^2{n^2}{h^2}}}$$
For electron,
Ee = $$ - {{{e^4}{m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}$$ = 13.6 eV
For muon,
Em = $$ - {{{e^4}\left( {207} \right){m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}$$
= 13.6 $$ \times $$ 207 = 2815.2 eV
In hydrogen atom one electron present. Now that the electron in hydrogen atom is replaced by a muon ($$\mu$$).
We know, Energy(E) = $$ - {{{e^4}m} \over {8\varepsilon _0^2{n^2}{h^2}}}$$
For electron,
Ee = $$ - {{{e^4}{m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}$$ = 13.6 eV
For muon,
Em = $$ - {{{e^4}\left( {207} \right){m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}$$
= 13.6 $$ \times $$ 207 = 2815.2 eV
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